A question about the equivalence principle.

Nick20

I had a physics test at school recently. One of the questions was based on the equivalence principle, going something like this: Two clocks in a space ship that is accelerating. One at the bottom and one at the top of the space ship. Now think that the space ship is so far away from any object in space, that it is not affected by any gravitational force.

It is my understanding that according to the equivalence principle, one can not be able to do any test that suggests that you are no longer in a gravitational field, but in an accelerated system. And according to the theory of relativity, the clock furthest down in the gravitational field will go slower than any clock higher up. Yet it makes no sense to me that the same rules would apply in an accelerated system. The correct answer supposedly is that the clock at the bottom of the ship will slow down. What am I missing? Would the clock at the bottom of the ship really slow down? Making the rules of time dilation apply in an accelerated system, the same way it does in an gravitational field?

A.T.

Yes. Time flows non-uniformly in non-inertial frames, like the rest frame of the ship.

DaleSpam

Yes. In fact, the fact that the clock "lower" in the accelerating system slows down can be derived purely using SR without any reference to the equivalence principle or the Einstein field equations.

The elegant way to do it is to determine the transform between the inertial and accelerating systems and then evaluate the spacetime interval for a stationary object.

The brute force way to do it is to calculate the Doppler shift between the upper and lower clocks.

yuiop

Consider the point of view of an inertial (non accelerating) observer who watches the rocket accelerating. If the length of the rocket appears constant to the accelerating observers on board the rocket, then the rocket appears to be length contracting according to the inertial observer. This means the back end of the rocket is accelerating faster than the front end and the back end of the rocket is moving faster than the front end at any given time according to that inertial observer, so the back end experiences more time dilation relative to the front end.

stevendaryl

The fact that the clock in the front of a rocket will tick faster than a clock in the rear of a rocket follows from SR alone. Here's a heuristic argument: From the point of view of the launch frame of the rocket, as the rocket accelerates, it becomes length-contracted. That means that the distance between the front of the rocket and the rear of the rocket decreases. That means that the rear of the rocket is traveling slightly faster than the front of the rocket. That means that time dilation affects the rear of the rocket more than the front of the rocket. That means that the rear clock runs slower than the front clock.

stevendaryl

The Doppler shift derivation is very nice, but there is something a little puzzling about it, which is this:

We can compute the discrepancy between the clock rates at the front and the rear of the rocket using SR time dilation and length contraction. But the Doppler shift derivation works whether or not the rocket undergoes length-contraction: Imagine, instead of a single rocket with two clocks, we have two identical rockets, one above the other. They launch identically, and the distance between the rockets remains constant (as measured in
the launch frame). Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second.

I'm just leaving this as a puzzle---I'm not saying the Doppler shift explanation is wrong, but it seems weird that it works whether or not there is length contraction, while the other explanation relies only on length contraction (and differential time dilation).

A.T.

But the frequency shift will be different than for a single rocket with constant proper length, right?

stevendaryl

Not immediately; the shifts will start out the same as in the case of a single rocket.

jartsa

I have always thought that when somebody carries the clock from the back end of the rocket to the front end of the rocket, that is when the clock experiences velocity time dilation.

But now I see that if I use the non-contracted lenght, then the time dilation is too large.
If I use the contracted length, then the time dilation is too small.

yuiop

This works because the velocity of the rocket at the rear at the time the signal is emitted, is lower than the velocity of the rocket at the front at the time the signal is received. This is actually a better explanation to the OP than the length contraction explanation, that largely ignores the relativity of simultaneity.

m4r35n357

Particularly since there should be no length contraction in the rocket frame, or have I missed the point badly here?

yuiop

I think you are along the right tracks. If there is no change of length in the rocket frame then the red shift observed in the rocket frame can only be explained by time dilation, because in that reference frame the ends of the rocket are stationary with respect to each other so there is no classical Doppler shift.

However, in the particular example we gave where the rocket is not length contracting in the inertial launch reference frame, the rocket will actually be length expanding in the rocket frame, but that is probably an unnecessary distraction and there will still be time dilation red shift over and above the classical Doppler shift.

P.S. I think you are also right to have concerns about the length contraction explanation (when no length contraction is measured in the rocket frame undergoing Born rigid acceleration) so that makes the explanation a bit weak. It does explain how the inertial observer in the launch frame concludes that the clocks at the back and front are running at different rates, and if we allow for the change in simultaneity when transforming to the rocket reference frame the conclusion still holds, but the explanation becomes more convoluted then while the Doppler explanation seems more direct. However, we still have to refer to the inertial reference frame to observe the different velocities of the back and front ends of the rocket, even for the Doppler explanation.

DaleSpam

I have actually never seen the length contraction argument worked out quantitatively, and I suspect that it will give the wrong value for exactly the reason that you mention. There is time dilation between two clocks that are undergoing identical accelerations in some inertial frame so that the time dilation is the same in that frame.

yuiop

The OP was asking how to demonstrate that the clock at the back of the ship runs slower than the clock at the front of the ship. Stevendaryl in his explanation that considers the case that has no length contraction in the inertial launch frame stated "Then even though the distance between the rockets remains constant, the Doppler shift explanation still holds: if the rear rocket sends signals at rate one per second up toward the front, the front will receive them at a rate slower than one per second." He seems to implying that we can conclude that the rear clock runs slower than the front clock from this observation. However, I think the reverse is true. If the front rocket sends signals at the rate of once per second, the rear will receive them at a rate slower than once per second. I may be wrong, as this is just my qualitative hunch. If that is true, then the red shift between the observers at the front at rear is reciprocal and it cannot be stated that the rear clock runs slower than the front clock with any certainty. The observed mutual redshift would be accounted for by the fact that they both see the other as going away from them (i.e. the observed redshift is mainly due to classical Newtonian Doppler shift.) If that is the case, then it would be necessary to invoke length contraction to explain the difference in clock rates at the front and rear.

Austin0

I think the signals received at the rear would be blue shifted. The acceleration between time of emission and reception would mean greater velocity at the reception point. In addition there would be greater time dilation by some non zero factor due to that increased velocity which would add blue shift. If you consider the ships separated by a great distance it is obvious there would be significant increase in the rear ships velocity relative to the ealrier velocity of emission.
It would seem to also apply in the other direction with the exception that in that case the difference in time dilation would cancel out part of the red shift due to the increased velocity
It also seems like there would be more red shift due to velocity measured in the front ship than blue shift due purely to velocity measured in the rear ship.
With the dilation factor working oppositely I guess it would take calculations for specific conditions to determine what the relationship would be.
Perhaps

yuiop

I have had a closer look and it looks like I was a bit hasty and you (and stevendaryl) are right. The rear observer does observe blue shift from signals coming from the leading rocket, even if there is no length contraction. This can be seen from examining the attached graph "f2b" which plots the paths (blue) of two rockets with equal and constant proper acceleration as seen in an inertial reference frame. The horizontal green lines are lines of equal proper time on the accelerating rockets at one unit time intervals. On this graph the diagonal black lines are signals sent from the front rocket to the rear rocket at regular unit time intervals. One interesting aspect is the the blue shift seen by the rear rocket reduces asymptotically towards unity as time progresses and the proper separation of the rockets increases.
This is correct too. The leading rocket does observe red shift of signals coming from the rear rocket with or without length contraction. The attached chart "b2f" charts the no length contraction case where the rear rocket sends signals at regular unit time intervals to the front rocket. Again there is an interesting aspect, in that the red shift increases dramatically over time, which is the exact opposite of how the blue shift seen by the rear rocket evolves.

The final attached chart is the length contraction case which is associated with the Rindler Metric and Born Rigid acceleration. In this case the distance between rockets and the blue or red shift from neighbouring rockets, remains constant over time, in an accelerating rocket reference frame. The green curves are the "lines of equal proper time" for the rocket and plotted at 1/2 unit time intervals. For those that might be interested, these curves are plotted using these parametric equations:

[tex]x = r * \cosh (T/r) [/tex]
[tex]t = r * sinh(T/r) [/tex]
where r is the parametric variable and T is a constant (The proper time). r is also the nominal radius and is inversely proportional to the constant proper acceleration of a given rocket.

Attached Thumbnails

     

Austin0

Hi WHy do you thnk there would be a difference between a single rocket and two rockets if the distance remained constant in the reference frame in both cases ?